Week 3bEE 42 and 100, Fall New topics – energy storage elements Capacitors Inductors

Week 3bEE 42 and 100, Fall Books on Reserve for EE 42 and 100 in the Bechtel Engineering Library “The Art of Electronics” by Horowitz and Hill (2 nd edition) -- A terrific source book on practical electronics “Electrical Engineering Uncovered” by White and Doering (2 nd edition) – Freshman intro to aspects of engineering and EE in particular ”Newton’s Telecom Dictionary: The authoritative resource for Telecommunications” by Newton (“18 th edition – he updates it annually) – A place to find definitions of all terms and acronyms connected with telecommunications. TK5102.N486 Shelved with dictionaries to right of entry gate.

Week 3bEE 42 and 100, Fall The Capacitor Two conductors (a,b) separated by an insulator: difference in potential = V ab => equal & opposite charges Q on conductors Q = CV ab where C is the capacitance of the structure,  positive (+) charge is on the conductor at higher potential Parallel-plate capacitor: area of the plates = A (m 2 ) separation between plates = d (m) dielectric permittivity of insulator =  (F/m) => capacitance (stored charge in terms of voltage) F(F) V ab + - +Q -Q

Week 3bEE 42 and 100, Fall Symbol: Units: Farads (Coulombs/Volt) Current-Voltage relationship: or Note: Q (v c ) must be a continuous function of time +vc–+vc– icic C C (typical range of values: 1 pF to 1  F; for “supercapa- citors” up to a few F!) Electrolytic (polarized) capacitor C If C (geometry) is unchanging, i C = dv C /dt

Week 3bEE 42 and 100, Fall Voltage in Terms of Current; Capacitor Uses Uses: Capacitors are used to store energy for camera flashbulbs, in filters that separate various frequency signals, and they appear as undesired “parasitic” elements in circuits where they usually degrade circuit performance

Week 3bEE 42 and 100, Fall 20056

Week 3bEE 42 and 100, Fall Schematic Symbol and Water Model for a Capacitor

Week 3bEE 42 and 100, Fall You might think the energy stored on a capacitor is QV = CV 2, which has the dimension of Joules. But during charging, the average voltage across the capacitor was only half the final value of V for a linear capacitor. Thus, energy is. Example: A 1 pF capacitance charged to 5 Volts has ½(5V) 2 (1pF) = 12.5 pJ (A 5F supercapacitor charged to 5 volts stores 63 J; if it discharged at a constant rate in 1 ms energy is discharged at a 63 kW rate!) Stored Energy CAPACITORS STORE ELECTRIC ENERGY

Week 3bEE 42 and 100, Fall vc–+vc– icic A more rigorous derivation

Week 3bEE 42 and 100, Fall Example: Current, Power & Energy for a Capacitor –+–+ v(t)v(t) 10  F i(t)i(t) t (  s) v (V) t (  s) i (  A) v c and q must be continuous functions of time; however, i c can be discontinuous. Note: In “steady state” (dc operation), time derivatives are zero  C is an open circuit

Week 3bEE 42 and 100, Fall w (J) –+–+ v(t)v(t) 10  F i(t)i(t) t (  s) p (W) t (  s)

Week 3bEE 42 and 100, Fall Capacitors in Parallel i(t)i(t) +v(t)–+v(t)– C1C1 C2C2 i1(t)i1(t)i2(t)i2(t) i(t)i(t) +v(t)–+v(t)– C eq Equivalent capacitance of capacitors in parallel is the sum

Week 3bEE 42 and 100, Fall Capacitors in Series i(t)i(t) C1C1 + v 1 (t) – i(t)i(t) + v(t)=v 1 (t)+v 2 (t) – C eq C2C2 + v 2 (t) – CCC eq 

Week 3bEE 42 and 100, Fall Capacitive Voltage Divider Q: Suppose the voltage applied across a series combination of capacitors is changed by  v. How will this affect the voltage across each individual capacitor? v+vv+v C1C1 C2C2 + v 2 (t)+  v 2 – + v 1 +  v 1 – +–+– Note that no net charge can can be introduced to this node. Therefore,  Q 1 +  Q 2 =0 Q1+Q1Q1+Q1 - Q 1  Q 1 Q2+Q2Q2+Q2  Q 2  Q 2 Q1=C1v1Q1=C1v1 Q2=C2v2Q2=C2v2 Note: Capacitors in series have the same incremental charge.

Week 3bEE 42 and 100, Fall MEMS Airbag Deployment Accelerometer Chip about 1 cm 2 holding in the middle an electromechanical accelerometer around which are electronic test and calibration circuits (Analog Devices, Inc.) Hundreds of millions have been sold. Airbag of car that crashed into the back of a stopped Mercedes. Within 0.3 seconds after deceleration the bag is supposed to be empty. Driver was not hurt in any way; chassis distortion meant that this car was written off.

Week 3bEE 42 and 100, Fall Application Example: MEMS Accelerometer to deploy the airbag in a vehicle collision Capacitive MEMS position sensor used to measure acceleration (by measuring force on a proof mass) MEMS = micro- electro-mechanical systems FIXED OUTER PLATES g1g1 g2g2

Week 3bEE 42 and 100, Fall Sensing the Differential Capacitance –Begin with capacitances electrically discharged –Fixed electrodes are then charged to +V s and –V s –Movable electrode (proof mass) is then charged to V o C1C1 C2C2 VsVs –Vs–Vs VoVo Circuit model

Week 3bEE 42 and 100, Fall Application: Condenser Microphone

Week 3bEE 42 and 100, Fall A capacitor can be constructed by interleaving the plates with two dielectric layers and rolling them up, to achieve a compact size. To achieve a small volume, a very thin dielectric with a high dielectric constant is desirable. However, dielectric materials break down and become conductors when the electric field (units: V/cm) is too high. –Real capacitors have maximum voltage ratings –An engineering trade-off exists between compact size and high voltage rating Practical Capacitors

Week 3bEE 42 and 100, Fall The Inductor An inductor is constructed by coiling a wire around some type of form. Current flowing through the coil creates a magnetic field and a magnetic flux that links the coil: Li L When the current changes, the magnetic flux changes  a voltage across the coil is induced: iLiL vL(t)vL(t) + _ Note: In “steady state” (dc operation), time derivatives are zero  L is a short circuit

Week 3bEE 42 and 100, Fall Symbol: Units: Henrys (Volts second / Ampere) Current in terms of voltage: Note: i L must be a continuous function of time +vL–+vL– iLiL L (typical range of values:  H to 10 H)

Week 3bEE 42 and 100, Fall Schematic Symbol and Water Model of an Inductor

Week 3bEE 42 and 100, Fall Stored Energy Consider an inductor having an initial current i(t 0 ) = i )( )()( )()()( 0 Li tw dptw titvtp t t      INDUCTORS STORE MAGNETIC ENERGY

Week 3bEE 42 and 100, Fall Inductors in Series v(t)v(t) L1L1 + v 1 (t) – v(t)v(t) + v(t)=v 1 (t)+v 2 (t) – L eq L2L2 + v 2 (t) – +–+– +–+– i(t)i(t) i(t)i(t) Equivalent inductance of inductors in series is the sum

Week 3bEE 42 and 100, Fall L1L1 i(t)i(t) i2i2 i1i1 Inductors in Parallel L2L2 +v(t)–+v(t)– L eq i(t)i(t) +v(t)–+v(t)–

Week 3bEE 42 and 100, Fall Capacitor v cannot change instantaneously i can change instantaneously Do not short-circuit a charged capacitor (-> infinite current!) n cap.’s in series: n cap.’s in parallel: Inductor i cannot change instantaneously v can change instantaneously Do not open-circuit an inductor with current (-> infinite voltage!) n ind.’s in series: n ind.’s in parallel: Summary

Week 3bEE 42 and 100, Fall Transformer – Two Coupled Inductors N 1 turns N 2 turns + - v1v1 + v2v2 - |v 2 |/|v 1 | = N 2 /N 1

Week 3bEE 42 and 100, Fall AC Power System

Week 3bEE 42 and 100, Fall High-Voltage Direct-Current Power Transmission Highest voltage +/- 600 kV, in Brazil – brings 50 Hz power from12,600 MW Itaipu hydropower plant to 60 Hz network in Sao Paulo

Week 3bEE 42 and 100, Fall Relative advantages of HVDC over HVAC power transmission Asynchronous interconnections (e.g., 50 Hz to 60 Hz system) Environmental – smaller footprint, can put in underground cables more economically,... Economical -- cheapest solution for long distances, smaller loss on same size of conductor (skin effect), terminal equipment cheaper Power flow control (bi-directional on same set of lines) Added benefits to the transmission (system stability, power quality, etc.)

Week 3bEE 42 and 100, Fall QuantityVariableUnitUnit Symbol Typical Values Defining Relations Important Equations Symbol ChargeQcoulombC1aC to 1Cmagnitude of × electron charges q e = x C i = dq/dt CurrentIampereA 1  A to 1kA 1A = 1C/s VoltageVvoltV 1  V to 500kV 1V = 1N-m/C Summary of Electrical Quantities

Week 3bEE 42 and 100, Fall PowerPwattW 1  W to 100MW 1W = 1J/sP = dU/dt; P=IV EnergyUjouleJ1fJ to 1TJ1J = 1N-mU = QV ForceFnewtonN 1N = 1kg- m/s 2 Timetseconds ResistanceRohm  1  to 10M  V = IR; P = V 2 /R = I 2 R R CapacitanceCfaradF1fF to 5FQ = CV; i = C(dv/dt);U = (1/2)CV 2 C InductanceLhenryH 1  H to 1H v = L(di/dt); U = (1/2)LI 2 L Summary of Electrical Quantities (concluded)